Question

Tìm m = 2^2010 - ( 2^2009 + 2^2008 + .... + 2^1 + 2^0 )

Answer 1

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(2^{2010}-M=1+2+2^2+...+2^{2008}+2^{2009}\) 

\(2\left(2^{2010}-M\right)=2+2^2+2^3+...+2^{2009}+2^{2010}\)

\(2\left(2^{2010}-M\right)-\left(2^{2010}-M\right)=\left(2+2^2+2^3+...+2^{2009}+2^{2010}\right)-\left(1+2+2^2+...+2^{2008}+2^{2009}\right)\)

\(2^{2010}-M=2^{2010}-1\)

\(M=2^{2010}-2^{2010}+1\)

\(M=1\)

Answer 2

Đặt \(M=2^{2010}-A\)

Ta có:

\(A=2^{2009}+2^{2008}+...+2^1+2^0\)

\(\Rightarrow2A=2^{2010}+2^{2009}+...+2^2+2^1\)

\(\Rightarrow2A-A=\left(2^{2010}+2^{2009}+...+2^2+2^1\right)-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(\Rightarrow A=2^{2010}-1\)

\(\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)\)

\(\Rightarrow M=\left(2^{2010}-2^{2010}\right)+1\)

\(\Rightarrow M=1\)